Can u help me to prove that the Z(Sym(n))={1}, centralizer of the symmetric group is equal to 1.

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- Jul 4th 2008, 07:01 AMdimukcentralizer of sym(n)
Can u help me to prove that the Z(Sym(n))={1}, centralizer of the symmetric group is equal to 1.

- Jul 4th 2008, 10:49 AMMatt WestwoodCentraliser of symmetric group order > 3 is trivial
The proof I've got starts like this ...

The symmetric group order $\displaystyle n$ can be considered as the set of all permutations of $\displaystyle n$ letters.

Let us choose a permutation $\displaystyle \tau \in Z (S_n), \tau \neq e$, so that $\displaystyle \tau (i) = j, i \neq j$.

Since $\displaystyle n \ge 3$, we can find another permutation $\displaystyle \rho$ which interchanges j and k $\displaystyle (k \ne i, j)$ and leaves everything else where it is. So the inverse of $\displaystyle \rho$ does the same thing, and both $\displaystyle \rho$ and $\displaystyle \rho^{-1}$ leave i alone.

Thus:

$\displaystyle

\rho \tau \rho^{-1} (i) = \rho \tau (i) = \rho (j) = k

$

So $\displaystyle \rho \tau \rho^{-1} (i) = k \ne j = \tau (i)$.

If $\displaystyle \rho$ and $\displaystyle \tau $ were to commute, $\displaystyle \rho \tau \rho^{-1} (i) = \tau (i)$ (right multiply by $\displaystyle \rho$). But they don't.

Whatever $\displaystyle \tau \in S_n$ is, you can always find a $\displaystyle \rho$ such that $\displaystyle \rho \tau \rho^{-1} \ne \tau $.

So no non-identity elements of $\displaystyle S_n$ commute with all elements of $\displaystyle S_n$. - Jul 4th 2008, 12:17 PMdimukcentralizer of Sym(n)
Thanx.

- Jul 7th 2008, 02:23 AMdimukcentralizer of Sym(n)
what do u mean by $\displaystyle \tau(i)$?

- Jul 7th 2008, 10:11 AMMatt Westwood
In this context, $\displaystyle S_n$ is the set of all permutations of $\displaystyle n$ objects (which we can call 1, 2, 3, ..., $\displaystyle n$ without loss of generality).

A permutation of $\displaystyle n$ objects is a rearrangement of them. For example, a rearrangement of (1, 2, 3) is (1, 3, 2), and another one is (2, 1, 3), and so on.

$\displaystyle \tau$ is an element of $\displaystyle S_n$, and so it is a rearrangement of $\displaystyle n$ objects.

So $\displaystyle \tau$ applied to one of the objects of an n-element set will either move it to somewhere else or leave it where it is.

For example, if $\displaystyle \tau$ is the permutation that changes (1, 2, 3) to (1, 3, 2), then $\displaystyle \tau (1) = 1, \tau (2) = 3$ and $\displaystyle \tau (3) = 2$.

In this context, $\displaystyle \tau$ is any permutation you care to think of, but not the identity permutation (which leaves everything where it is). So there's bound to be at least one element of this n-element set which is not left where it is.

Call this element $\displaystyle i$.

What I mean by $\displaystyle \tau(i)$ is the result of applying this arbitrarily selected permutation $\displaystyle \tau$ to an equally arbitrarily selected element $\displaystyle i$, and calling it $\displaystyle j$.

Don't confuse $\displaystyle S_n$ (the set of permutations on $\displaystyle n$ obects) with the actual objects that $\displaystyle S_n$ is acting on. They are not the same thing. - Jul 7th 2008, 11:31 AMdimukcentralizer of Sym(n)
thanx matt.